Linear First Order ODE/y' = x + y

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Theorem

The linear first order ODE:

$(1): \quad \dfrac {\d y} {\d x} = x + y$

has the general solution:

$y = C e^x - x - 1$


Initial Value: $\map y 0 = 1$

The linear first order ODE:

$(1): \quad \dfrac {\d y} {\d x} = x + y$

with initial condition:

$\map y 0 = 1$

has the particular solution:

$y = 2 e^x - x - 1$


Proof

Rearranging $(1)$:

$(2): \quad \dfrac {\d y} {\d x} - y = x$

$(2)$ is a linear first order ODE in the form:

$\dfrac {\d y} {\d x} + \map P x y = \map Q x$

where:

$\map P x = -1$
$\map Q x = x$


Thus:

\(\displaystyle \int \map P x \rd x\) \(=\) \(\displaystyle -\int \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle - x\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle e^{\int P \rd x}\) \(=\) \(\displaystyle e^{-x}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {e^x}\)


Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:

\(\displaystyle \map {\dfrac {\d} {\d x} } {\dfrac y {e^x} }\) \(=\) \(\displaystyle \frac x {e^x}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \dfrac y {e^x}\) \(=\) \(\displaystyle \int \frac x {e^x} \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle -\frac 1 {e^x} \paren {x + 1} + C\) Primitive of $x e^{a x}$
\(\displaystyle \leadsto \ \ \) \(\displaystyle y\) \(=\) \(\displaystyle C e^x - x - 1\)

$\blacksquare$


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